Answer:
6.67 M
Explanation:
Molarity =
<em>Convert 200g NaOH to moles. Convert 750 mL to L.</em>
200 g NaOH x (1 mol/39.998 g) = 5.00025... mol NaOH
750 mL x (1 L/1000 mL) = 0.750 L
<em>Substitute values into the equation.</em>
Molarity =
Molarity = 6.667... M
Molarity = 6.67 M
<u>61.25 grams</u> of CO can be formed from 35 grams of oxygen.
The molecular mass of oxygen is <u>16 gmol⁻¹</u>
The molecular mass of carbon monoxide is<u> 28 gmol⁻¹</u>
Explanation:
The molar mass of carbon monoxide is molar mass of C added to that of O;
12 + 16 = 28
= 28g/mol
The molar mass of oxygen is 16 g/mol while that of oxygen gas (O₂) is 32 g/mol
Since the ration oxygen to carbon monoxide is 1: 2 moles, we begin to find out how many moles of carbon monoxide are formed by 35 g of oxygen;
35/32 * 2
= 70/32 moles
Then multiply by the molar mass of carbon monoxide;
70/32 * 28
= 61.25 g
Answer is: mass of unused sulfur is 5.87 grams.
Balanced chemical reaction: C + 2S → CS₂.
m(C) = 12.0 g; mass of carbon.
m(S) = 70.0 g; mass of sulfur.
n(C) = m(C) ÷ M(C).
n(C) = 12 g ÷ 12 g/mol.
n(C) = 1 mol; amount of substance.
n(S) = m(S) ÷ M(S).
n(S) = 70 g ÷ 32.065 g/mol.
n(S) = 2.183 mol.
From chemical reaction: n(C) : n₁(S) = 1 : 2.
n₁(S) = 1 mol · 2 = 2 mol.
Δn(S) = n(S) - n₁(S).
Δn(S) = 2.183 mol - 2 mol.
Δn(S) = 0.183 mol; amount of unused sulfur.
Δm(S) = 0.183 mol · 32.065 g/mol.
Δm(S) = 5.87 g.
Answer:
pH is 7.60
Explanation:
Let's think the equations:
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
As every weak acid, we make an equilbrium
The salt is dissociated in solution
NaClO → Na⁺ + ClO⁻
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
Initially 0.3m 0.35m
We have the moles of acid, and the moles of conjugate base.
Reacts x X X
Some amount has reacted, so I obtained (in equilibrium) the moles of base + that amount, and the same amount for H₃O⁺ (ratio is 1:1)
HClO + H₂O ⇄ ClO⁻ + H₃O⁺
0.3 - x 0.35 + x x
Let's make the expression for Ka
Ka = [ClO⁻] . [H₃O⁺] / [HClO]
(we don't add water, because it is included in Ka)
2.9x10⁻⁸ = (0.35+x).x / (0.3-x)
Ka is in order of 10⁻⁸, I can assume that 0.3-x is 0.3 and 0.35 +x =0.3
2.9x10⁻⁸ = (0.35)x / (0.3)
(2.9x10⁻⁸ . 0.3) /0.35 = x
2.48x10⁻⁸ = x
This is [H₃O⁺]
For pH = - log [H₃O⁺]
pH = 7.60
The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.